Question -
Answer -
(i) 16x2 – 9y2 + 32x + 36y – 164= 0
Given:
The equation => 16x2 – 9y2 +32x + 36y – 164 = 0
Let us find the centre, eccentricity, foci and directions ofthe hyperbola
By using the given equation
16x2 – 9y2 + 32x + 36y – 164= 0
16x2 + 32x + 16 – 9y2 + 36y– 36 – 16 + 36 – 164 = 0
16(x2 + 2x + 1) – 9(y2 – 4y+ 4) – 16 + 36 – 164 = 0
16(x2 + 2x + 1) – 9(y2 – 4y+ 4) – 144 = 0
16(x + 1)2 – 9(y – 2)2 = 144
Here, center of the hyperbola is (-1, 2)
So, let x + 1 = X and y – 2 = Y
The obtained equation is of the form
Where, a = 3 and b = 4
Eccentricity is given by:
Foci: The coordinates of the foci are (±ae, 0)
X = ±5 and Y = 0
x + 1 = ±5 and y – 2 = 0
x = ±5 – 1 and y = 2
x = 4, -6 and y = 2
So, Foci: (4, 2) (-6, 2)
Equation of directrix are:
∴ The center is (-1, 2), eccentricity (e) = 5/3, Foci = (4,2) (-6, 2), Equation of directrix = 5x – 4 = 0 and 5x + 14 = 0
(ii) x2 – y2 + 4x = 0
Given:
The equation => x2 – y2 +4x = 0
Let us find the centre, eccentricity, foci and directions ofthe hyperbola
By using the given equation
x2 – y2 + 4x = 0
x2 + 4x + 4 – y2 – 4 = 0
(x + 2)2 – y2 = 4
Here, center of the hyperbola is (2, 0)
So, let x – 2 = X
The obtained equation is of the form
Where, a = 2 and b = 2
Eccentricity is given by:
Foci: The coordinates of the foci are (±ae, 0)
X = ± 2√2and Y = 0
X + 2 = ± 2√2and Y = 0
X= ± 2√2– 2 and Y = 0
So, Foci = (± 2√2– 2, 0)
Equation of directrix are:
∴ The center is (-2, 0), eccentricity (e) = √2, Foci = (-2± 2√2, 0), Equation of directrix=
x + 2 = ±√2
(iii) x2 – 3y2 – 2x = 8
Given:
The equation => x2 – 3y2 –2x = 8
Let us find the centre, eccentricity, foci and directions ofthe hyperbola
By using the given equation
x2 – 3y2 – 2x = 8
x2 – 2x + 1 – 3y2 – 1 = 8
(x – 1)2 – 3y2 = 9
Here, center of the hyperbola is (1, 0)
So, let x – 1 = X
The obtained equation is of the form
Where, a = 3 and b = √3
Eccentricity is given by:
Foci: The coordinates of the foci are (±ae, 0)
X = ± 2√3and Y = 0
X – 1 = ± 2√3and Y = 0
X= ± 2√3+ 1 and Y = 0
So, Foci = (1 ± 2√3,0)
Equation of directrix are:
∴ The center is (1, 0), eccentricity (e) = 2√3/3, Foci =(1 ± 2√3, 0), Equationof directrix =
X = 1±9/2√3